# Question: Rudin exercise 3.8

***If $$\sum{a_n}$$ converges and $$\{b_n\}$$ is bounded and monotonic, prove that $$\sum{a_nb_n}$$ converges. ***

This is my work thus far: I have $$\sum_na_n$$ is convergent which implies for all $$\varepsilon > 0$$ there exists $$N$$ s.t.

$$\left\vert \sum_{k=n}^ma_k\right\vert =|s_m-s_n|\leq \varepsilon$$ for $$m \geq n\geq N$$ where $$\{s_n\}$$ is a Cauchy sequence.

$$\{b_n\}$$ is bounded so there exists $$M\in \mathbb{N}$$ s.t. $$\vert b_n \vert \leq M.$$

Then we have $$\vert a_nb_n\vert=|a_n|b_n|\leq |a_n|M.$$

I know $$\left\vert \sum_{n}a_k\right\vert\leq \sum_n|a_n|$$ in general but I don't have that $$\sum_na_n$$ converges absolutely. I haven't learned Dirichlet's Test yet.

What I want to do is show that $$|a_nb_n|\leq c_n, n\leq N_0\in \mathbb{N}$$ and $$\sum c_n$$ converges. For now, I think my $$c_n=|a_n|M$$ but I don't see how I can show $$\sum_n |a_n|M=M\sum_n |a_n|$$ converges. Any tips please?

• You can not show that $M\sum |a_n|$ converges. It would implies that $\sum a_n$ converges absolutely. But this is not a hypothesis. Nov 12, 2020 at 17:05
• Agreed! Do you think the answer I posted below resolves my problem? Nov 12, 2020 at 17:11
• I do not understand the following step $|\sum a_kb_k|\leq |\sum a_k M |$ Nov 12, 2020 at 17:14
• @PAM1499, would it make sense if I just remove that in the chain of equalities? I wanted the M to be in my equation so I could use it to bound $|\sum a_nb_n|$ to prove convergence. Nov 12, 2020 at 19:58
• Just to clarify why what you did is not right define $a_0=-\frac{1}{3}$, $a_n=\frac{(-1)^{n+1}}{2^n}$ for $n \geq 1$ then $\sum a_n=0$. Choose $b_0=0$ and $b_n=1$ for $n \geq 1$. Then $\{b_n\}$ is non-decreasing and bounded by $M=1$. But $|\sum a_nb_n|=\frac{1}{3} >0=|\sum a_n M|$. Nov 12, 2020 at 20:12

$$\{b_n\}$$ is bounded so $$|b_n|\leq M$$.
$$\sum_n a_n$$ is convergent so $$\forall \varepsilon >0, \exists N$$ s.t. $$\vert \sum_n^m a_k|\leq \varepsilon/M,\,\,m\leq n \leq N.$$
$$\vert \sum_{k=n}^m a_kb_k\vert\leq \vert \sum_{k=n}^m a_k M|=|M\sum_{k=n}^m a_k\vert=M\vert\sum_{k=n}^m a_k\vert\leq M\cdot \varepsilon/M=\varepsilon.$$